Question

You throw a ball at a height of 6 feet above the

ground. The height h (in feet) of the ball after t seconds can be modeled by the equation

h=-16t² +62t +6. After how many seconds does the ball reach a height of 27 feet?

Answer 1

0.375 second and 3.5 second

Explanation:

The position can be modeled by a quadratic function
`\displaystyle{h=-16t^2+62t+6}`. We are tasked to find the time when a ball reaches a height of 27 feet. Therefore, let h = 27:

`\displaystyle{27=-16t^2+62t+6}`

Solve for t:

`\displaystyle{27-6=-16t^2+62t}\\\\\displaystyle{21=-16t^2+62t}\\\\\displaystyle{16t^2-62t+21=0}`

Since the equation is quite complicated and more time-consuming to solve, i'll skip the factoring or quadratic part:

`\displaystyle{t=0.375, 3.5}`

After done solving the equation, you'll get t = 0.375 and 3.5 seconds. These solutions are valid since both are positive values and time can only be positive.

Hence, it'll take 0.375 and 3.5 seconds for a ball to reach 27 feet.